Weekly challenge 29: Integral equation

7064 /www/nrich/html/content/id/7064/ 1 2011-03-15T10:31:10 <mdoxml version="1.0"> <ul id="buttonBar"> <li> <a href="http://nrich.maths.org/6552&part=">Warm-up problem</a> </li> <li> <a href="http://nrich.maths.org/4934&part=">Try this next</a> </li> <li> <a href="https://nrich.maths.org/discus/messages/27/27.html">Ask NRICH</a> </li> <li> <a href="http://nrich.maths.org/1365&part=">Read all about it</a> </li> <li> <a href="http://nrich.maths.org/6495&part=solution">Last week's solution</a> </li> </ul> <div> Find the function $f(x)$ which solves the equation $$\int_0^x f(t)\,dt = 3f(x)+k\,,$$ where $k$ is a constant. </div> <div class="framework"> Did you know ... ? In the same way that a differential equation is formed from differentials, an integral equation is formed from integrals. Problems in mathematics might naturally be specified in terms of integrals, others in terms of differentials. Differentials are mainly used when the problem involves only local changes, such as the force at a point, whereas integrals are used where the problem involves a global property of a system, such as the total energy.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The integral equation is: $$\int_0^x f(t)\,dt = 3f(x)+k,\quad \quad(\star)$$ where $k$ is a constant. Differentiating both sides of $(\star)$ gives $$f(x) = 3f'(x)$$ If there is a solution of $(\star)$ it must be of the form $$f(x) = Ae^{x/3},$$ for some constant $A$. We check to see whether or not this is a solution. For $f(x)=Ae^{x/3}$ we have $$\int_0^x Ae^{t/3}\,dt = \Big[3Ae^{t/3}\Big]_0^x = 3Ae^{x/3}-3A.$$ Thus $f(x)=Ae^{x/3}$ is a solution if and only if $A=-k/3$. The unique solution is $$f(x) = {-k\over 3} e^{x/3}.$$ It was also noted by Nat that the problem can be solved rapidly using the university-level technique of Laplace transforms. He says You could have it done with the Laplace transform (which I think is more ideal if you really want to solve integro-diff equations). So let $s$ be the frequential variable and $F(s)$ the Laplace transform of $f(x)$ Then $$\frac{F(s)}{s} =3F(s) + \frac{k}{s}$$ $$F(s) = - \frac{k}{3}\cdot \frac{1}{s-\frac{1}{3}}$$ Then $$f(x) = - \frac{k}{3}\cdot e^{\frac{x}{3}}$$ </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The integral equation is: $$\int_0^x f(t)\,dt = 3f(x)+k,\quad \quad(\star)$$ where $k$ is a constant. Differentiating both sides of $(\star)$ gives $$f(x) = 3f'(x)$$ If there is a solution of $(\star)$ it must be of the form $$f(x) = Ae^{x/3},$$ for some constant $A$. We check to see whether or not this is a solution. For $f(x)=Ae^{x/3}$ we have $$\int_0^x Ae^{t/3}\,dt = \Big[3Ae^{t/3}\Big]_0^x = 3Ae^{x/3}-3A.$$ Thus $f(x)=Ae^{x/3}$ is a solution if and only if $A=-k/3$. The unique solution is $$f(x) = {-k\over 3} e^{x/3}.$$ </mdoxml> 2 3 0 0 0 0 1 Weekly challenge 29: Integral equation A weekly challenge: these are shorter problems aimed at Post-16 students or enthusiastic younger students. Collections Weekly Challenge